Notes
Graphs
$f(x) = ax^2 + bx + c$ is a parabola. It opens up if $a > 0$ and down if $a < 0$. The axis of symmetry is $\displaystyle x = -\frac{b}{2a}$, and the vertex is on it. Vertex form $a(x - h)^2 + k$ shows the vertex $(h, k)$.
Four ways to solve $ax^2 + bx + c = 0$
| Method | Use it when |
|---|---|
| Factoring | the trinomial factors nicely |
| Square roots | there is no $x$ term, like $x^2 = 20$ |
| Completing the square | $a = 1$ and $b$ is even, or you need vertex form |
| Quadratic formula | always works: $\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ |
The discriminant $b^2 - 4ac$ tells you how many real solutions: positive gives 2, zero gives 1, negative gives 0. The real solutions are the graph's $x$-intercepts.
Choosing a model
Linear: constant differences. Quadratic: constant second differences. Exponential: constant ratios.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$(x - 4)(x + 2) = 0$.
The vertex is $(3, 7)$ and the parabola opens down.
$x^2 = 25$.
$1 - 24 = -23 < 0$.
$-16(t^2 - 2t - 3) = 0 \Rightarrow (t - 3)(t + 1) = 0$; time can't be negative.
Complete the square: $(x - 3)^2 = 7$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A rocket is launched, and its height in meters after $t$ seconds is $h(t) = -5t^2 + 40t$.
(a) When does the rocket reach its maximum height, and what is that height?
Show the answer and scoring guide
$\displaystyle t = -\frac{40}{2(-5)} = 4$ seconds. $h(4) = -80 + 160 = 80$ meters.
Check off the points you earned:
(b) When does the rocket land? Show your work.
Show the answer and scoring guide
$-5t^2 + 40t = 0 \Rightarrow -5t(t - 8) = 0$, so $t = 0$ (launch) or $t = 8$. It lands after 8 seconds.
Check off the points you earned:
(c) Between which times is the rocket higher than 60 meters?
Show the answer and scoring guide
$-5t^2 + 40t = 60 \Rightarrow t^2 - 8t + 12 = 0 \Rightarrow (t - 2)(t - 6) = 0$. It is above 60 meters between 2 and 6 seconds.
Check off the points you earned: